A circle is circumscribed around triangle ABC, find length?

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Helly123
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IMG_0936_1.jpg


some formula related
nratbig.gif


I tried to draw the problem
Untitled.png


can anyone give me clue how to solve it?
 
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Helly123 said:
... can anyone give me clue how to solve it?
Hi Helly123

In order to be helped you must first tell us what is asked. Also, think about it for a while and tell how do you think you must go about solving it (at least in rough lines).
 
Arc-length is ##l = r \theta##.
 
QuantumQuest said:
Hi Helly123

In order to be helped you must first tell us what is asked. Also, think about it for a while and tell how do you think you must go about solving it (at least in rough lines).
the question : (1) how the length of arcs AB, BC, CA, expressed in a, b, c, r.
(2) I don't know what kind of expression in number 2
(3) find length of AB, BC, CA, when a = 75 degrees, b = 60, c = 45, r = 1. without trigonometric function
upload_2017-6-13_6-48-14.png
 
Buffu said:
Arc-length is ##l = r \theta##.
can you give me more clue...
 
Helly123 said:
can you give me more clue...

Can you express arclengths in terms of ##a,b,c ,r ## ?
 
Buffu said:
Can you express arclengths in terms of ##a,b,c ,r ## ?
maybe like
AB / 2phi R = c / 360 degrees
AB = c/360 . 2 phi R
 
Helly123 said:
maybe like
AB / 2phi R = c / 360 degrees
AB = c/360 . 2 phi R

No I did not get what you did. I guess finding arclengths confused you, can you find the angle subtended by each side at centre ?

Also in the question what ##\bbox[5px,Border: 2px solid black]{2-2}## mean ? what does that number mean ? I have never seen such a weird way to denote a unknown.
 
Buffu said:
No I did not get what you did. I guess finding arclengths confused you, can you find the angle subtended by each side at centre ?

Also in the question what ##\bbox[5px,Border: 2px solid black]{2-2}## mean ? what does that number mean ? I have never seen such a weird way to denote a unknown.
Maybe the way i draw the circle and triangle is wrong?
 
Buffu said:
No I did not get what you did. I guess finding arclengths confused you, can you find the angle subtended by each side at centre ?

Also in the question what ##\bbox[5px,Border: 2px solid black]{2-2}## mean ? what does that number mean ? I have never seen such a weird way to denote a unknown.

That's, the first time i saw it too...

" can you find the angle subtended by each side at centre ? " how to find that?
 
g4220-6-2-28asd.png


Find theta.
 
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cnh1995 said:
Looks good.

Your image link contains a lot of ads and the actual image might take some time to load.
Next time onwards, use the upload button to directly add the images in the post.
Ok thanks, I am using phone so i sent link. Usually i upload. (Cant upload through phone) btw sir, my answer is right? And the formula too?
 
Helly123 said:
my answer is right? And the formula too?
Yes.
Helly123 said:
Cant upload through phone) b
I can upload through phone using the upload button below the 'post reply' button.

Helly123 said:
btw sir,
I'm no sir..just a guy like you, hanging out here!
 
Helly123 said:
Ok thanks, I am using phone so i sent link. Usually i upload. (Cant upload through phone) btw sir, my answer is right? And the formula too?

Now you need to find the area of each small triangle AOC ...
 
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cnh1995 said:
Yes.

I can upload through phone using the upload button below the 'post reply' button.
I know that button. I tried before in another thread. But the image wouldn't show up
 
Buffu said:
Now you need to find the area of each small triangle AOC ...
Ok, for the area of AOC = 1/2 r.r.sin 2b
Area CAB = 1/2 r.r sin 2a
Area BCA = 1/2 r.r sin 2c
So the entire triangle is 1/2. r^2 ( sin 2a + sin 2b + sin 2c)

3) ... AB = 90degrees.1cm ?
BC = 150 degrees.1cm?
AC = 120 degrees.1cm
 
Helly123 said:
Ok, for the area of AOC = 1/2 r.r.sin 2b
Area CAB = 1/2 r.r sin 2a
Area BCA = 1/2 r.r sin 2c
So the entire triangle is 1/2. r^2 ( sin 2a + sin 2b + sin 2c)

3) ... AB = 90degrees.1cm ?
BC = 150 degrees.1cm?
AC = 120 degrees.1cm

2 is correct.

In three the angles should be in radians.
 
AC = 2/3 phi rad
BC = 5/6 phi rad
AB = 1/2 phi rad

The key answer is root 2, (root 6 + root 2)/2 , root 3
 
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Why my answer not match the key answer? :(
 
Helly123 said:
Why my answer not match the key answer? :(

The question wants length of sides of the triangle not the length of the arcs which you calculated.
 
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The boxes with 2-1, 2-2, 2-3, etc in them seem to number the subquestions. Kind of where you would fill in the answers to question 2-1, 2-2, 2-3, etc. Except you probably have to fill them in on a separate answer sheet.
 
Hendrik Boom said:
The boxes with 2-1, 2-2, 2-3, etc in them seem to number the subquestions. Kind of where you would fill in the answers to question 2-1, 2-2, 2-3, etc. Except you probably have to fill them in on a separate answer sheet.
I think so...
 
Buffu said:
Sir, i want to know how to find arc AB BC CA if the triangle like what I draw...
untitled-png.png
 
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Helly123 said:
Sir, i want to know how to find arc AB BC CA if the triangle like what I draw... View attachment 205419

It is same process.
Where are you stuck ?
 
Buffu said:
It is same process.
Where are you stuck ?
I meant, totally everything stuck..
what's the correlation of let's say angle ABC to R, since ABC not on the same center as circle center?
and L = tetha . R cannot be applied anymore. since the R not in the triangle.
 
Helly123 said:
I meant, totally everything stuck..
what's the correlation of let's say angle ABC to R, since ABC not on the same center as circle center?
and L = tetha . R cannot be applied anymore. since the R not in the triangle.

Ok first try with angle subtended on the centre by the line AB.

Also ##l = r\theta## is applicable here since it does not matter whether it is in te triangle or not. It is true by the definition of angle in radians.

I think what you are confused with is the fact that if angle b is half the angle subtended at the centre by the line AB or not. It is also true.
 
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Buffu said:
Ok first try with angle subtended on the centre by the line AB.

Also ##l = r\theta## is applicable here since it does not matter whether it is in te triangle or not. It is true by the definition of angle in radians.

I think what you are confused with is the fact that if angle b is half the angle subtended at the centre by the line AB or not. It is also true.

the angle substended by line AB is the c angle?

And even though the triangle is like what i have, the tetha substended by line AB twice angle c,
Buffu said:
Ok first try with angle subtended on the centre by the line AB.

Also ##l = r\theta## is applicable here since it does not matter whether it is in te triangle or not. It is true by the definition of angle in radians.

I think what you are confused with is the fact that if angle b is half the angle subtended at the centre by the line AB or not. It is also true.
if I draw tetha in my triangle like this, and want to find arc AC, but ended up finding arc AX
2_-_2015_mat_a.png
if I draw this, the length AX and CX no longer R... because the center not in X point

how is it?
2_-_2015_mat_a_2.png
 
Helly123 said:
the angle substended by line AB is the c angle?

And even though the triangle is like what i have, the tetha substended by line AB twice angle c,
if I draw tetha in my triangle like this, and want to find arc AC, but ended up finding arc AX
View attachment 205447if I draw this, the length AX and CX no longer R... because the center not in X point

how is it?
View attachment 205448
Yes sorry it is angle c.

Your figure should be like this :-
g4220-6-2-2dsf8.png
 
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